Convergence of Characteristic Functions and Distributions

In this appendix I will provide the proof of the univariate version of Theorem 6.22. Let Fn be a sequence of distribution functions on К with corresponding characteristic functions yn (t), and let F be a distribution function on К with characteristic function y(t) = limn—TO^n(t). Let

F(x) = limliminf Fn (x + 8 ), F (x) = lim limsup Fn (x + 8 ).

5^0 n——to 5^0 n — TO

The function F(x) is right continuous and monotonic nondecreasing in x but not necessarily a distribution function itself because limx^TO F(x) may be less than 1 or even 0. On the other hand, it is easy to verify that limx;—TO F(x) = 0. Therefore, if limx—TO F(x) = 1, then F is a distribution function. The same applies to F(x): If limx—TO F(x) = 1, then F is a distribution function.

I will first show that limxF(x) = limx^O F(x) = 1 and then that F(x) =

F(x).

Lemma 6.C.1: Let Fn be a sequence of distribution functions on К with corre­sponding characteristic functions pn (t) and suppose that p(t) = limn^Opn (t) pointwise for each t in R where p is continuous in t = 0. Then F(x) = lim5.[0Hminfn^OFn (x + 5) is a distribution function and so is F(x) = Hms.[olimsupn^O Fn (x + 5).

where fzn is the probability measure on the Borel sets in К corresponding to Fn. Hence, if we put T = A-1 it follows from (6.75) that

which can be rewritten as

Now let 2A and -2A be continuity points of F. Then it follows from (6.77), the condition that v(t) = limn^x Vn(t) pointwise for each t in R, and the bounded9 convergence theorem that

Because v(0) = 1 and v is continuous in 0 the integral in (6.78) converges to 2 for A ^x.

Moreover, F(-2 A) I 0 if A ^ x. Consequently, it follows from (6.78) that limAxxF(2A) = 1. By the same argument it follows that limAxxF(2A) = 1. Thus, F and F are distribution functions. Q. E.D.

Lemma 6.C.2: Let Fn be a sequence of distribution functions on R such that F(x) = lims.[0 liminfnxx Fn(x + S)and F(x) = lims^0 limsupnxxFn(x + 8) are distribution functions. Then for every bounded continuous function v on R and every є > 0 there exist subsequences nk(є) and n k(є) such that

Proof: Without loss of generality we may assume that v(x) є [0, 1] for all x. For any є > 0 we can choose continuity points a < b of F(x) such that F(b) – F(a) > 1 – є. Moreover, we can choose continuity points a = c1 <

Let p*(t) be the characteristic functionof F. Because p (t) = limn—TO pn (t),it follows from Lemma 6.C.2 that for each t and arbitrary є > 0, |p(t) – pjyt)| < є; hence, p(t) = p*(t). The same result holds for the characteristic function p*(t) of F : p(t) = p*(t). Consequently, p(t) = limn—TO pn(t) is the character­istic function of both F and F, whichbyLemma6.C.1 are distribution functions. By the uniqueness of characteristic functions (see Appendix 2.C in Chapter 2) it follows that both distributions are equal: F(x) = F(x) = F(x), for instance. Thus, for each continuity point x of F, F(x) = limn—TOFn(x).

Note that we have not assumed from the outset that p(t) = limn—TOpn (t) is a characteristic function but only that this pointwise limit exists and is continuous in zero. Consequently, the univariate version of the “if” part of Theorem 6.22 can be restated more generally as follows: